How do you solve #4x - (x * 3^(1/2)) = 6#?
1 Answer
Explanation:
The idea hee is to isolate
To do that, start by using
#4x - x * 3^(1/2) = 6#
#x * (4 - 3^(1/2)) = 6#
You can rewrite the equation by replacing the fractional exponent by its corresponding radical term
#x * (4 -sqrt(3)) = 6#
DIvide both sides of the equation by
#( x* color(red)(cancel(color(black)(4-sqrt(3)))))/(color(red)(cancel(color(black)(4-sqrt(3))))) = 6/(4 -sqrt(3))#
#x = 6/(4 - sqrt(3))#
Rationalize the denominator of the fraction by multiplying it by
#6/(4 - sqrt(3)) * (4 + sqrt(3))/(4 + sqrt(3)) = (6 * (4 + sqrt(3)))/((4 - sqrt(3))(4 + sqrt(3))#
#= (6 * (4 + sqrt(3)))/(4^2 - (sqrt(3))^2)#
#= 6/13 * (4 + sqrt(3))#
Therefore,
#x = color(green)(6/13 * (4 + sqrt(3)))#